Question: Evaluate $\int^{\pi/2}_03\cos^3x\,dx\,$.
Since there is an odd power of $~\cos x\,$, we first factor out a $~\cos x~$ and use the identity $~\sin^2x+\cos^2x=1~$ to rewrite the other factor of $~\cos^2x\,$. $ 3\int^{\pi/2}_0\cos^3x\, dx = 3\int^{\pi/2}_0\cos x\cdot\cos^2x\,dx$ $ ~~\,=3\int^{\pi/2}_0\cos x\,(1-\sin^2x)\,dx$ We now use a $~u$ -substitution with $ u=\sin x~~~~~$ and $~~~~~du=\cos x\,dx\,$. It is appropriate to change the limits of integration as well. $ x=0 \Rightarrow u=0$ $ x=\dfrac\pi2 \Rightarrow u=1$ Then we have a new integration problem. $ 3\int^{\pi/2}_0\cos x\,(1-\sin^2x)\,dx=3\int_0^1(1-u^2)\,du\,$. Evaluating the definite integral is now straightforward. $ 3\int_0^1(1-u^2)\,du= \Big(3u-u^3\Big)\Bigg|_0^1$ $ ~~\,=\Big(3-1\Big)-\Big(0-0\Big)=2$